Algebraic Field Extensions and Irreducible Polynomials

Let $$E/K$$ be a field extension, $$a,b\in E$$ algebraic over $$K$$. Show: $$\text{Min}(a,K,X)$$ irreducible over $$K(b)$$ if and only if $$\text{Min}(b,K,X)$$ irreducible over $$K(a)$$

My attemp:

(1) Let deg $$\text{Min}(a,K,X)$$ = n, deg $$\text{Min}(b,K,X)$$ = m. We know: $$[F(ab):F] = [F(ab):F(a)]\cdot [F(a):F] = [F(ab):F(a)]\cdot n$$ $$[F(ab):F] = [F(ab):F(b)]\cdot [F(b):F] = [F(ab):F(b)]\cdot m$$

$$\Rightarrow [F(ab):F(a)]\cdot n = [F(ab):F(b)]\cdot m$$

Let $$f:= \text{Min}(a,K,X)$$ irreducible over $$K(b)$$. Since f is minimal Polynomial of $$a$$ over $$K[X]$$, it is the unique normalized irreducible polynomial with $$f(a)=0$$. Let $$f'(a)=0$$ and $$f'(X)\in F(b)[X]$$ be the minimal polynomial of $$a$$ over $$K(b)$$. Since $$f\in K[X]$$, also $$f\in K(b)[X]$$ and $$\text{deg}\, f'\leq \text{deg}\, f$$. Because $$f:= \text{Min}(a,K,X)$$ irreducible over $$K(b) \Rightarrow f'=f \Rightarrow \text{deg}\, f'= \text{deg}\, f\Rightarrow [F(ab):F(b)]=n$$.

Is this ok?

• you switch from $$K$$ to $$F$$ and back (that's of course no serious problem)
• you want to write $$F(a, b)$$ (or $$K(a, b)$$) instead of $$F(ab)$$
Also note that if you use the arguments of this answer, you'll see that your conditions are both equivalent to the symmetric condition $$K(a) \cap K(b) = K$$.